Vince went on a $3$ day hiking trip. Each day, he walked $\dfrac34$ the distance that he walked the day before. He walked $83.25$ kilometers total in the trip. How far did Vince walk on the $1^\text{st}$ day of the trip? Round your final answer to the nearest kilometer.
Solution: Notice that Vince's daily distances form a geometric sequence. The total distance Vince hikes after $ n$ days is the ${\text{sum}}$ of the first $n$ terms in the sequence. This is called a geometric series. This is the formula for that sum: $ S={a}\left(\dfrac{1-{r}^{ n}}{1-{r}}\right)$ where ${a}$ is the first term and ${r}$ is the common ratio. We can use this formula, along with the given information, to find the distance Vince walked on the $1^\text{st}$ day of the trip, $ a$. Using the given information We are given that each daily distance is ${\dfrac34}$ of the previous day's distance. This is the common ratio $ r$. We are given that the total distance after $ 3$ days is ${83.25\text{ km}}$. So the sum of series $ S$ is ${83.25\text{ km}}$, and the number of terms $ n$ is $ 3$. We are looking for the distance Vince walked on the $1^{\text{st}}$ day, $ a$. Finding the first term $\begin{aligned} {83.25}&={a} \cdot \dfrac{1-\left({0.75}\right)^{{3}}}{1-\left({0.75}\right)} \\\\ \dfrac{1-\left({0.75}\right)}{1-\left({0.75}\right)^{{3}}} \cdot {83.25} &= {a} \\\\ 36 &= {a} \end{aligned}$ Answer Vince walked $36\,\text{km}$ on the $1^\text{st}$ day of the trip.